Statistics Test

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Statistics Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

If we increase the mean of the given data by adding 3 to each observation, then by what percentage would the mean increase when compared to the original mean?

3, 5, 9, 12, 5, 7, 3, 4 and 6

A
150%

B
100%

C
40%

D
50%

Solution

Given,
3, 5, 9, 12, 5, 7, 3, 4 and 6 are 9 observations.

_{Mean of 9 observations =}

=
= = 6

So when 3 is added to each observation, the mean would also be increased by 3 and for 9 observations, 3 × 9 = 27 will be added to the sum.
Therefore, new mean = 9
Percentage of the new mean to the previous mean can be calculated as:

× 100 = 150%

Percentage increase of the new mean to that of the previous mean = 150% - 100% = 50%
Hence, option 4 is the answer.
Question 2
1 / -0

If the mode of the given data is 2, then what would be the mean of the given data?

2, x, 4, 2, 3, 4, 5, 7, 9, 2, 7, 4

A
2.25

B
3.67

C
4.25

D
5.36

Solution

Given,
The mode of 2, x, 4, 2, 3, 4, 5, 7, 9, 2, 7, 4 is 2.
We know that,
Mode = Maximum number of repeated values in the data.
In this data, 2 and 4 are repeated values and are repeated three times each.
But as the mode is 2, it is clear that 'x' should be 2 so as to get our mode as 2.
Therefore, the sum of all the values = 2 + 2 + 4 + 2 + 3 + 4 + 5 + 7 + 9 + 2 + 7 + 4 = 51
Total number of observations = 12

_{Mean =}
Therefore, mean = = 4.25

Hence, option 2 is the answer.
Question 3
1 / -0

If the median of p + 2, p + 3, p + 8, p + 4 and p + 5 is 12, then what would be the difference between the median and the mean of the data (Where p is a positive number)?

A
0.8

B
0.6

C
0.4

D
0.3

Solution

Given,
The median of p + 2, p + 3, p + 8, p + 4 and p + 5 is 12.
p is a positive number.
So, we have to arrange the data in the ascending order, i.e. p + 2, p + 3, p + 4, p + 5 and p + 8.
As there are 5 observations, the median will be:
= 3^rd term i.e. p + 4
Given median = 12
Therefore, p + 4 = 12
p = 8
Now put the value of p in all the observations.
P + 2 = 10
p + 3 = 11
p + 4 = 12
p + 5 = 13
and p + 8 = 16
_{Mean =}

Now, mean = = 12.4

Therefore, the difference between the mean and the median of the data = 12.4 – 12 = 0.4
Hence, option 3 is the answer.
Question 4
1 / -0

Mean of 30 observations was 56 when one number, that is 42 was misread as 24. If the mistake is corrected and each observation is multiplied by 3, what will be the new mean?

A
170

B
169.8

C
165.2

D
159.5

Solution

Given, mean of 30 observations was 56 when one number 42 was misread as 24.
Therefore, the sum of all the 30 observations is = (56 × 30) = 1680
But as one of the values was misunderstood, 18 (42 - 24) must be added to the sum in order to find the new mean.
Therefore, new sum = 1680 + 18 = 1698
Now, multiply new sum by 3 = 5094
Hence, new mean = = 169.8
Hence, 169.8 is the answer.
Question 5
1 / -0

If the mode is 18, what would be the ratio of the missing value and the median of the data?

11, 12, 8, 10, 16, 18, 24, 21, 18, x, 11, 31, 43, 36, 39, 41, 42, 28, 45, 46 and 30.

A
4 : 3

B
3 : 4

C
5 : 6

D
2 : 3

Solution

Given observations are:
11, 12, 8, 10, 16, 18, 24, 21, 18, 11, 31, 43, 36, 39, 41, 42, 28, 45, 46 and 30.
and 'x' has no fixed position.
As mode = 18, i.e. maximum number of times repeated value.
Therefore, x = 18 as all the values are written once except for 11 and if x = 18, then 18 would be repeated thrice and hence will be the mode.
The data is arranged in the ascending order.
8, 10, 11, 11, 12,16, 18, 18, 18, 21, 24, 28, 30, 31, 36, 39, 41, 42, 43, 45 and 46.
Hence, there are 21 values in total and the median is the middle value of the data which equals 24.
Now, the ratio of the missing value and the median of the data = 18 : 24
= 3 : 4
Hence, option 2 is the answer.
Question 6
1 / -0

Find the value of x from the following scores (out of 10), scored by 30 kabaddi players in the series of matches where mean is 10 and median is 9:

5, 5, 6, 7, 9, 5, 4, 5, 6, 5, 9, 5, 6, 5, 7, 1, 2, 4, 7, 3, 2, x, 7, 7, 4, 3, 1, 0, 7.

A
x = 7

B
x = 5

C
x = 4

D
x = 2

Solution

Given : Mean = 10 and median = 9
Mode = 3 median – 2 mean
= 3 × 9 – 2 × 10
= 27 – 20 = 7
Therefore, Mode = 7
x = 7 {Mode, which means the number which appears most often in a set of numbers, is 7.}
Question 7
1 / -0

If the average time taken by 12 students for making project A is 5 hours and the average time taken by 8 students for making project B is 3 hours, then what would be the mean time of the students for both the projects?

A
4.2 hours

B
5.8 hours

C
6.8 hours

D
7.1 hours

Solution

Given,
Average time of 12 students for making project A = 5 hours
Total time = 12 × 5 = 60 hours
Average time of 8 students for making project B = 3 hours
Total time = 8 × 3 = 24 hours
Total time taken = (60 + 24) hours = 84 hours
Now, mean time for both the projects = = 4.2 hours
Question 8
1 / -0

In a data of 50 observations, the mean of the first 35 observations is 53 and the last 15 observations is 25. If 5 is added to each of the first 35 observations and 3 is subtracted from each of the last 15 observations, then what is the mean of the 50 observations altogether?

A
47.2

B
45

C
40

D
39.5

Solution

Mean of 35 observations after adding 5 = 53 + 5 = 58
New sum of 35 observations = 58 × 35 = 2030
Similarly, mean of 15 observations after subtracting 3 = 25 - 3 = 22
New sum of 15 observations = 22 × 15 = 330
Now, total sum of 50 observations = 2030 + 330 = 2360
New mean == 47.2
Question 9
1 / -0

The students in a school are arranged in a queue in descending order as per their heights. If the median of heights (in cm) of students are 197, 195, 192, 187, 185, 183, 181, 180, p, 178, 174, 172, 169, 162, 150, 149, 145, q is 179, where p is double in height as compared to q, then what is height (in cm) of q?

A
90

B
80

C
60

D
50

Solution

Here number of terms (n) = 18 (even)

Median will be the average of (n/2)^th term and {(n/2) +1}^thterms

Median = average of term and
(+ 1)^th term
Median = average of 9^th and 10^th term
179 =(178 + p)
179 × 2 = 178 + p
358 – 178 = p
p = 180
Then, q =[p is double of q]
q = 90
Question 10
1 / -0

Given that p is the mean of a₁, a₂, ... , a_n. If we add z (not equal to zero) in each observation and after that multiply each observation with z, what will be the new mean?

A
zp + z²

B
p(z + 1)

C

D

Solution

If z is added to each observation, then mean would also increase by z and become p + z. Then if each observation is multiplied by z, the mean would also become z times, i.e. z(p + z) = zp + z²

Question 11

1 / -0

If the mean of the following marks is 30, then what will be the number of students who scored the highest marks?

Marks	Number of students
10	2
15	5
20	10
25	A
35	9
45	A + 2

16

14

15

25 Solution


10	2	20
15	5	75
20	10	200
25	A	25A
35	9	315
45	A + 2	45A + 90
Total	2A + 28	70A + 700

Mean =

30 =

60A + 840 = 70A + 700
840 – 700 = 70A – 60A
140 = 10A

A =

A = 14
Therefore, total number of students who scored 45 marks, which is the highest score = 14 + 2 = 16

Question 12
1 / -0

In a class there are 40 students and their mean marks are 80. If 20 students who have mean marks equal to 15, leave the class and after that 15 more student who have mean marks 30, admit in the class, find the new mean (approximate).

A
95.71

B
105.71

C
125.52

D
150.36

Solution

Mean marks of 40 students = 80
Sum of marks of 40 students = 80 × 40 = 3200
Now, mean marks of 20 students who leave = 15
Sum of marks of 20 students who leave = 15 × 20 = 300
Now, mean marks of 15 students who take admission = 30
Sum of marks of 15 students who take admission = 15 × 30 = 450
Total students = 40 – 20 + 15 = 35
Sum of marks of 35 students = 3200 – 300 + 450 = 3350
New mean =
= 95.71 (approximately)
Question 13
1 / -0

Find the difference between the median and the mean of all the two-digit numbers in which the ones digit is one more than the tens digit.

A
20

B
0

C
10

D
0.5

Solution

Given,
The two-digit numbers are such that the second digit is the consecutive number after the first digit.
Therefore, the numbers would be:
12, 23, 34, 45, 56, 67, 78, 89
Now, as the numbers are already arranged in the ascending order and there are 8 numbers in total,

Median =

= = 50.5

Now, Mean =

= = 50.5

Now, the difference between the median and the mean of these two-digit numbers = 50.5 – 50.5 = 0
Hence, 0 is the answer.
Question 14
1 / -0

Which of the following terms in the given series must not be changed in order to keep the median same?

78, 23, 56, 1, 22, 89, 101, 103

A
101 and 103

B
1 and 103

C
22 and 23

D
56 and 78
Solution
The median is the middle of a sorted list of numbers.
If we have 'n' terms, median when number of observations is:
- Even: Average of ^th term and ^th term.
- Odd: ()^th term
If we arrange the numbers in ascending order; 1, 22, 23, 56, 78, 89, 101, 103 (Total 8 terms).
The median is the average of the 4^th and 5^th terms.
So, these terms should not be changed to keep the median same.

Question 15

1 / -0

Directions: This question is based on the table below:

Number of shares	Price of each share ($)
10	15
15	18
12	12
23	5

What is the median price of shares?

5

12

15

18 Solution

Number of shares = 60
Median =
= (30^th + 31^st) = (12 + 12) = 12

Question 16

1 / -0

The frequencies of numbers 3, 5, 6 and 4 are x, x + 2, x - 8 and x, respectively. If their mean is 4, then the value of x will be

3

6

7

8 Solution

x_i	f_i	x_if_i
3	x	3x
5	x + 2	5x + 10
6	x - 8	6x - 48
4	x	4x
	4x - 6	18x - 38

Now, according to the question:
= 4
18x - 38 = 16x - 24
2x = - 24 + 38
2x = 14
x = 7

Question 17

1 / -0

A shop-keeper made a list of the weights of the cakes (in grams) sold by him in a day, which is as follows:
200, 250, 500, 200, 500, 1000, 550, 750, 200, 1000, 950, 900, 850, 750, 650, 650, 500, 200, 350, 450, 400, 500, 550, 750, 500, 850, 950, 1000
If the intervals in the data are maintained in such a way: 1 - 200, 201 - 400, 401 - 600, 601 - 800, 801 - 1000, find the ratio of the frequency of the number of cakes sold in category 201 - 400 to 401 - 600.

5 : 7

5 : 3

4 : 7

3 : 8

Solution

Weights of the cakes sold by the shop-keeper: 200, 250, 500, 200, 500, 1000, 550, 750, 200, 1000, 950, 900, 850, 750, 650, 650, 500, 200, 350, 450, 400, 500, 550, 750, 500, 850, 950, 1000

After sorting this data we have, 200, 200, 200, 200, 250, 350, 400, 450, 500, 500, 500, 500, 500, 550, 550, 650, 650, 750, 750, 750, 850, 850, 900, 950, 950, 1000, 1000, 1000

From given data this table is formed:

Mass (in grams)	frequency
1 - 200	4
201 - 400	3
401 - 600	8
601 - 800	5
801 - 1 000	8

From the above table it is clear that the ratio required = 201 - 400 : 401 - 600 = 3 : 8 = 3 : 8

Question 18
1 / -0

The mean of x + a, x + 2, x + 4, x + 13 and 2(x + 5) is 16. Given, 10 > a > 4, 10 > x > 4 and (x + 4) < (x + a) < (x + 13). The median of the given data is the same as that of the mean.

Find the value of 'a'.

A
7

B
9

C
10

D
11

Solution

Given, the mean of x + a, x + 2, x + 4, x + 13 and 2(x + 5) is 16.
Number of observations = 5

Mean =
Mean of the given observations =
=

But the given mean = 16

Therefore, 16 =
80 = 6x + a + 29
6x + a = 80 – 29 = 51
6x + a = 51 … Eq 1
Now, x < 10 > a
Also, here (x + 4) < (x + a) < (x + 13)
Therefore, it is clear that the smallest value would be (x + 2).
Also (x + 4) < (x + a) < (x + 13) and at last 2(x + 5) for 10 > x > 4 and 10 > a > 4.
Therefore, the order is:
(x + 2) < (x + 4) < (x + a) < (x + 13) < (2 x + 10)
Here it is clear that the data is arranged in the ascending order and as number of observations is 5,
Median = = 3^rd term.
As median is the same as mean = 16
16 = (x + a) (Because 3^rd term) … Eq 2
Subtracting equation 2 from equation 1 we get:
5 x = 35
x = 7
As x + a = 16
7 + a = 16
a = 16 – 7 = 9
Hence, option 2 is the answer.
Question 19
1 / -0

In a departmental store on Wednesday, 15 candies, 11 chocolates, and 15 t-shirts were sold. On the same day, there was a sale of denims and jackets where the sale of denims was double the sale of jackets. If the average count of all the products sold was 16, find the number of denims and jackets sold together.

A
38

B
39

C
40

D
41

Solution

According to question, average count of all the products sold on Wednesday = 16
Number of different products available = 5
So, total goods sold = 16 × 5 = 80
Let the number of jackets sold be x.
So, the number of denims sold = 2x (given)
According to the given instructions,
15 candies + 11 chocolates + 15 t-shirts + 2x denims + x jackets = 80
So, 15 + 11 + 15 + 2x + x = 80
41 + 3x = 80
3x = 80 – 41 = 39
x = 39 ÷ 3 = 13
Number of jackets = x = 13
Number of denims = 2x = 2 × 13 = 26
So, number of denims and jackets sold together = 26 + 13 = 39
Question 20
1 / -0

In a park there are 120 marbles in which red, blue and green marbles are in the ratio 5 : 4 : 3. The average weight of the blue marbles is 30 grams, that of the red marbles is 60 grams and that of the green marbles is half that of the blue marbles. Find the mean weight of all the marbles.

A
38.75 grams

B
39.75 grams

C
40.75 grams

D
41.75 grams

Solution

Total marbles = 120
Ratio of all type of marbles = Red : Blue : Green = 5 : 4 : 3
Mean weight = (60 × 5 + 30 × 4 + 15 × 3) ÷ 12
= (300 + 120 + 45) ÷ 12 = 38.75 grams

Question 21

1 / -0

Directions: Read the following frequency table carefully and answer the question.

X_i	(f_i)
150	15
300	13
450	9
600	p
750	6
900	9
1050	6
1200	5
Total	63 + p

If the mean is , find the value of p.

8

7

5

6 Solution

X_i	(f_i)	f_ix_i
150	15	2250
300	13	3900
450	9	4050
600	p	600p
750	6	4500
900	9	8100
1050	6	6300
1200	5	6000
Total	63 + p	35100 + 600 p

Mean =

=

Solving this we get,

p = 7

Question 22
1 / -0

The arithmetic mean of 7n and 9 is a, that of 8n and 38 is b and that of 6n and 28 is c. According to this information, what would the arithmetic mean of a, b and c be?

A
(7n + 25)

B
5(7n +16)

C
(9n + 10)

D
(20n + 7)

Solution

a =

b =

c =
Arithmetic mean of a, b, c =

=
=
= (7n + 25)

Question 23

1 / -0

Match the following:

Column - I	Column - II
(I) The frequency of a certain set of observations can be found by using the _____.	(P) Median
(II) In data arranged in ascending order, _____ will be the value of observations which will divide the data into two equal groups.	(Q) Mode
(III) In a class of 30 students, the highest marks in a Science test are 78 and the lowest marks are 13. ______of the marks of these students will be 65.	(R) Tally marks
(IV) In a class of 20 students, the weight of 16 students is approximately 30 kg each which represents the _____ of 20 students.	(S) Range

(I) (P)

(II) (Q)

(III) (R)

(IV) (S)

(I) (P)

(II) (Q)

(III) (S)

(IV) (R)

(I) (R)

(II) (P)

(III) (S)

(IV) (Q)

(I) (S)

(II) (Q)

(III) (R)

(IV) (P)

Solution

(I) The frequency of a certain data can be found by using the tally marks.
(II) A median is that value of the observation which divides the data into two equal parts when the data is arranged in ascending order.
(III) Range = Highest value – Lowest value = 78 – 13 = 65
(IV) The observation that occurs most frequently, i.e. the observation with maximum frequency is called mode.
So, 30 kg is the mode.

Question 24

1 / -0

State 'T' for true and 'F' for false for the following statements:
(1) The mean of a certain set of observations is. If each observation is multiplied by

, the new value of the mean will be.
( 2) The scores of A are 35 which can be represented in tally as

.
(3) The following data has been arranged in ascending order. If the median of the data is 9, the value of x is 4.
8, 2x, x + 6, 25
(4) The mean of 7, 5, 18, 24, n + 4 and n is 13. The value of n is 10.
(5) The marks of 30 students of a class in a Science test are as follows:

32	23	18	33	42	11	24	29	34	5
49	23	22	44	45	28	22	44	30	33
12	45	45	33	48	34	35	49	26	13

The frequency of the students who obtained the highest marks is 3.

1 2 3 4 5

T T F F F

1 2 3 4 5

F F T F F

1 2 3 4 5

F F T T F

1 2 3 4 5

F T F T F

Solution

(1) Mean =
New mean =×
= (False)
If each observation is multiplied by 1/4 then the mean would also be reduced to one quarter.

(2) His marks are

= 6 × 5 = 30 (False)
(3) Median =
9 =
18 = 3x + 6
12 = 3x
x = 4 (True)

(4) Mean == 13
= 13
58 + 2n = 78
2n = 78 – 58
n = 10 (True)

(5) Highest marks = 49
Frequency = Number of times 49 occurs in the given data = 2 (False)

Question 25

1 / -0

Fill in the blanks.

(1) If the highest points scored by a person in a game are 5 more than the lowest marks scored by him, the range of this data is ____A____.
(2) If the frequencies of a class interval are a, 4a and 3a + 1, then the cumulative frequency of this class interval is ____B___.
(3) If the mean and the median of a certain set of observations are 4m and 6n respectively, then the mode is __C___.
(4) If the ratio of the mean and the median of a certain set of observations is 5 : 7, then the ratio of its mode to mean will be ____D____.

A 5

B (8a + 1)

C 2(9n - 4m)

D 11 : 5

A 4x

B 7a

C 2(3n - 2m)

D Upper limit

A 5 - x

B 3a + 1

C 2(9n + 4m)

D Mid value

A 5x

B a

C 2(3n + 2m)

D Range

Solution

(A) Let lowest scores = x
Highest scores = 5 + x
Range = Highest marks – Lowest marks
= 5 + x – x
= 5
The range of a set of data is the difference between the largest and smallest values.

(B) Cumulative frequency of a class interval is obtained by adding all the frequencies up to that class interval.
So, a + 4a + 3a + 1
= 8a + 1

(C) Mode = 3Median – 2Mean
Median = 6n, Mean = 4m
Mode = (6n × 3) - ( 2 × 4m)
= 18n – 8m
= 2(9n – 4m)

(D) Mean : Median = 5 : 7
Let,
Mean = 5x
Median = 7x
Mode = 3Median – 2Mean
= 3 × 7x – 2 × 5x
= 21x – 10x
= 11x
Ratio = 11x : 5x
= 11 : 5

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Statistics Test - 6

Result

Statistics Test - 6

Score

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Rank

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SHARING IS CARING

Question 1

150%

100%

40%

50%

Solution

Question 2

2.25

3.67

4.25

5.36

Solution

Question 3

0.8

0.6

0.4

0.3

Solution

Question 4

170

169.8

165.2

159.5

Solution

Question 5

4 : 3

3 : 4

5 : 6

2 : 3

Solution

Question 6

x = 7

x = 5

x = 4

x = 2

Solution

Question 7

4.2 hours

5.8 hours

6.8 hours

7.1 hours

Solution

Question 8

47.2

45

40

39.5

Solution

Question 9

90

80

60

50

Solution

Question 10

zp + z2

p(z + 1)

Solution

Question 11

16

14

15

25

Solution

Question 12

95.71

105.71

125.52

150.36

Solution

Question 13

20

zp + z²

(I) (P)

(II) (Q)

(III) (R)

(IV) (S)

(I) (P)

(II) (Q)

(III) (S)

(IV) (R)

(I) (R)

(II) (P)

(III) (S)

(IV) (Q)

(I) (S)

(II) (Q)

(III) (R)

(IV) (P)

1 2 3 4 5

T T F F F

1 2 3 4 5

F F T F F

1 2 3 4 5

F F T T F